Two real numbers are chosen. The first number belongs to the interval $[-2, 2]$, and the second number is positive and does not exceed 4. We want to determine the probability that the second number is less than the square of the first number.

1. Problem Description

Two real numbers are chosen. The first number belongs to the interval $[-2, 2]$, and the second number is positive and does not exceed

4. We want to determine the probability that the second number is less than the square of the first number.

2. Solution Steps

Let $x$ be the first number and $y$ be the second number.

We have $-2 \le x \le 2$ and $0 < y \le 4$.

We want to find the probability that $y < x^2$.

The total area of the region is a rectangle with sides of length $2 - (-2) = 4$ and $4 - 0 = 4$, so the total area is $4 \times 4 = 16$.

We need to find the area of the region where $y < x^2$ and $-2 \le x \le 2$ and $0 < y \le 4$.

The region of interest is bounded by $y=x^2$, $y=4$, $x=-2$ and $x=2$.

The area of this region is given by

$A = \int_{-2}^{2} \min(x^2, 4) dx$

Since $x^2 \le 4$ for $-2 \le x \le 2$, we have

$A = \int_{-2}^{2} x^2 dx = \left[ \frac{x^3}{3} \right]_{-2}^{2} = \frac{2^3}{3} - \frac{(-2)^3}{3} = \frac{8}{3} - \frac{-8}{3} = \frac{16}{3}$

The probability is the ratio of the area where $y < x^2$ to the total area.

$P = \frac{A}{\text{Total Area}} = \frac{16/3}{16} = \frac{16}{3 \times 16} = \frac{1}{3}$

3. Final Answer

The probability that the second number is less than the square of the first number is $\frac{1}{3}$.